But that is so obvious that even WM should have been able to figure it out on his own,

Every natural is, by definition, the determiner of that FISON of which it is the largest member, so that every natural is automatically in at least one FISON. > >> > Also. from a previous post:> >> > WM's claim:> >> > > If *not* all naturals in one s, then> > > exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.

That he has not done so suggests that he cannot do so because there are no such numbers satisfying his claimed condition.> > that is obvious in mathematics if two conditions have to hold:> 1) All naturals in FISONs.> 2) Not all naturals in one and the same FISONs.

If it were so obvious, even someone so inept at proofs as WM should be able to demonstrate it, but as he cannot, or at least has not, it would almost certainly be a false claim.

But it is well known that the natural numbers satisfy the trichotomy properties of a linearly ordered set which disproves WWM's claim.

And, at least everywhere outside of Wolkenmuekenheim,1) All naturals are in FISONs.2) Not all naturals are in one and the same FISONs.3) NOT exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.--